GATE Electrical Engineering 2022

Question 1
The transfer function of a real system, H(s), is given as:
H(s)=\frac{As+B}{s^2+Cs+D}
where A, B, C and D are positive constants. This system cannot operate as
A
low pass filter.
B
high pass filter
C
band pass filter.
D
an integrator.
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 1 Explanation: 
Put s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}
So, the system pass low frequency component. Put s=\infty , H(\infty )=0
For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.
Question 2
For an ideal MOSFET biased in saturation, the magnitude of the small signal current gain for a common drain amplifier is
A
0
B
1
C
100
D
infinite
Analog Electronics   Small Signal Analysis
Question 2 Explanation: 
For ideal MOSFET, i_G=0
Therefore, Current gain, A_I=\frac{i_s}{i_G}=\infty
Question 3
The most commonly used relay, for the protection of an alternator against loss of excitation, is
A
offset Mho relay.
B
over current relay.
C
differential relay
D
Buchholz relay.
Power Systems   Switch Gear and Protection
Question 4
The geometric mean radius of a conductor, having four equal strands with each strand of radius 'r', as shown in the figure below, is

A
4r
B
1.414r
C
2r
D
1.723r
Power Systems   Performance of Transmission Lines, Line Parameters and Corona
Question 4 Explanation: 
Redraw the configuration:

\therefore \; GMR=(r' \times 2r\times 2r\times 2\sqrt{2}r)^{1/4}
Where, r'=0.7788r
Hence, GMR=1.723r
Question 5
The valid positive, negative and zero sequence impedances (in p.u.), respectively, for a 220 kV, fully transposed three-phase transmission line, from the given choices are
A
1.1, 0.15 and 0.08
B
0.15, 0.15 and 0.35
C
0.2, 0.2 and 0.2
D
0.1, 0.3 and 0.1
Power Systems   Fault Analysis
Question 5 Explanation: 
We have,
X_0 \gt X_1=X_2
(for 3-\phi transposed transmission line)
Question 6
The steady state output (V_{out}), of the circuit shown below, will

A
saturate to +V_{DD}
B
saturate to -V_{EE}
C
become equal to 0.1 V
D
become equal to -0.1 V
Analog Electronics   Operational Amplifiers
Question 6 Explanation: 
Redraw the circuit:

From circuit,
\begin{aligned} V_{out} &=-\frac{1}{C_1}\int I\cdot dt \\ &= -\frac{1}{R_1C_1}\int 0 \cdot 1dt \\ \\ &=-\frac{0.1}{R_1C_1}\int dt \\ \\ &= -\frac{0.1}{R_1C_1}t \end{aligned}


Hence, V_{out}=-V_{EE}
Question 7
The Bode magnitude plot of a first order stable system is constant with frequency. The asymptotic value of the high frequency phase, for the system, is -180^{\circ}. This system has

A
one LHP pole and one RHP zero at the same frequency
B
one LHP pole and one LHP zero at the same frequency
C
two LHP poles and one RHP zero
D
two RHP poles and one LHP zero.
Control Systems   Frequency Response Analysis
Question 7 Explanation: 
The given system is non-minimum phase system Therefore, transfer function, T.F=\frac{s-1}{s+1}
Hence, one LHP pole and one RHP zero at the same frequency.
Question 8
A balanced Wheatstone bridge ABCD has the following arm resistances:
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
A
30\Omega \pm 3\Omega
B
30\Omega \pm 0.9\Omega
C
3000\Omega \pm 90\Omega
D
3000\Omega \pm 3\Omega
Electrical and Electronic Measurements   A.C. Bridges
Question 8 Explanation: 
The condition for balanced bridge
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
Question 9
The open loop transfer function of a unity gain negative feedback system is given by G(s)=\frac{k}{s^2+4s-5}.
The range of k for which the system is stable, is
A
k \gt 3
B
k \lt 3
C
k \gt 5
D
k \lt 5
Control Systems   Root Locus Techniques
Question 9 Explanation: 
Characteristic equation:
\begin{aligned} 1+G(s)H(s)&=0\\ 1+\frac{k}{s^2+4s-5}&=0\\ s^2+4s+k-5&=0 \end{aligned}
R-H criteria:
\left.\begin{matrix} s^2\\ s^1\\ s^0 \end{matrix}\right| \begin{matrix} 1 & k-5\\ 4 & 0\\ k-5 & \end{matrix}
Hence, for stable system,
k-5 \gt 0 \;\; \Rightarrow \; k \gt 5
Question 10
Consider a 3 x 3 matrix A whose (i,j)-th element, a_{i,j}=(i-j)^3. Then the matrix A will be
A
symmetric.
B
skew-symmetric.
C
unitary
D
null.
Engineering Mathematics   Linear Algebra
Question 10 Explanation: 
for \; i=j\Rightarrow a_{ij}=(i-i)^3=0\forall i
for \; i\neq j\Rightarrow a_{ij}=(i-j)^3=(-(j-i))^3=-(j-i)^3=-a_{ji}
\therefore \; A_{3 \times 3 } is skew symmetric matrix.
There are 10 questions to complete.

GATE Electrical Engineering 2021

Question 1
Let p and q be real numbers such that p^{2}+q^{2}=1. The eigenvalues of the matrix \begin{bmatrix} p & q\\ q& -p \end{bmatrix} are
A
1 and 1
B
1 and -1
C
j and -j
D
pq and -pq
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
Question 2
Let p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3, where z is a complex number.
Which one of the following is true?
A
\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right ) for all z
B
The sum of the roots of p\left ( z \right )=0 is a real number
C
The complex roots of the equation p\left ( z \right )=0 come in conjugate pairs
D
All the roots cannot be real
Engineering Mathematics   Complex Variables
Question 2 Explanation: 
Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.
Question 3
Let f\left ( x \right ) be a real-valued function such that {f}'\left ( x_{0} \right )=0 for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0 for all x \in \left ( 0,1 \right ). Then f\left ( x \right ) has
A
no local minimum in (0,1)
B
one local maximum in (0,1)
C
exactly one local minimum in (0,1)
D
two distinct local minima in (0,1)
Engineering Mathematics   Calculus
Question 3 Explanation: 
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 4
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

A
\text{10 V} in series with 12\:\Omega
B
\text{65 V} in series with 15\:\Omega
C
\text{50 V} in series with 2\:\Omega
D
\text{35 V} in series with 2\:\Omega
Electric Circuits   Network Theorems
Question 4 Explanation: 
Given circuit can be resolved as shown below,


V_{T H}=15+50=65 \mathrm{~V}


\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
Question 5
Which one of the following vector functions represents a magnetic field \overrightarrow{B}?
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
A
10x\hat{X}+20y\hat{Y}-30z\hat{Z}
B
10y\hat{X}+20x\hat{Y}-10z\hat{Z}
C
10z\hat{X}+20y\hat{Y}-30x\hat{Z}
D
10x\hat{X}-30z\hat{Y}+20y\hat{Z}
Electromagnetic Theory   Magnetostatic Fields
Question 5 Explanation: 
If \vec{B} is magnetic flux density then \vec{\nabla} \cdot \vec{B}=0
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
Question 6
If the input x(t) and output y(t) of a system are related as y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right ), then the system is
A
linear and time-variant
B
linear and time-invariant
C
non-linear and time-variant
D
non-linear and time-invariant
Signals and Systems   Linear Time Invariant Systems
Question 6 Explanation: 
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}


Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1


\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Question 7
Two discrete-time linear time-invariant systems with impulse responses h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ] and h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ] are connected in cascade, where \delta \left [ n\right ] is the Kronecker delta. The impulse response of the cascaded system is
A
\delta \left [ n-2\right ]+\delta \left [ n+1 \right ]
B
\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ]
C
\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ]
D
\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ]
Signals and Systems   Introduction of C.T. and D.T. Signals
Question 7 Explanation: 
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
Question 8
Consider the table given:
\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}
The correct combination that relates the constructional feature, machine type and mitigation is
A
P-V-X, Q-U-Z, R-T-Y
B
P-U-X, Q-S-Y, R-V-Z
C
P-T-Y, Q-V-Z, R-S-X
D
P-U-X, Q-V-Y, R-T-Z
Electrical Machines   Synchronous Machines
Question 8 Explanation: 
P: Damper bars used in synchronous machine (U) to prevent hunting (X)
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
Question 9
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is N_{L}. The balanced Newton-Raphson method is used to carry out load flow study in polar form. \text{H, S, M, and R} are sub-matrices of the Jacobian matrix J as shown below:
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
A
N_{L}\times \left ( N-1 \right )
B
\left ( N-1 \right )\times \left ( N-1-N_{L} \right )
C
N_{L}\times \left ( N-1+N_{L} \right )
D
\left ( N-1 \right )\times \left ( N-1+N_{L} \right )
Power Systems   Load Flow Studies
Question 9 Explanation: 
\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
Question 10
Two generators have cost functions F_{1} and F_{2}. Their incremental-cost characteristics are
\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}
\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}
They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are
A
P_{1}=P_{2}=130
B
P_{1}=160, P_{2}=100
C
P_{1}=140, P_{2}=120
D
P_{1}=120, P_{2}=140
Power Systems   Economic Power Generation and Load Dispatch
Question 10 Explanation: 
\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}
Solving equation (i) and (ii),
P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}
There are 10 questions to complete.

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Power Electronics Miscellaneous

Question 1
A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. L_{par} is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.
A
1
B
2
C
3
D
4
GATE EE 2020   Power Electronics
Question 1 Explanation: 


Using KCL, I_s=I_L-I_D
For inductively loaded circuits, load can be assumed to be constant.
\therefore \; I_s is maximum when, I_D is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.
Question 2
For the circuit shown in the figure below, assume that diodes D_1,D_2 and D_3 are ideal.

The DC components of voltages v_1 \; and \; v_2, respectively are
A
0 V and 1 V
B
-0.5 V and 0.5 V
C
1 V and 0.5 V
D
1 V and 1 V
GATE EE 2017-SET-1   Power Electronics
Question 2 Explanation: 


\begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\\ &= -0.5V \end{aligned}
Question 3
Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: V_{1}=500 kV, V_{2}=485 kV and V_{3}=1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible?
A
V_{1}=-500 kV,V_{2}=-485 kV and latex]I=1.5kA[/latex]
B
V_{1}=-485 kV,V_{2}=500 kV and latex]I=1.5kA[/latex]
C
V_{1}=500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex]
D
V_{1}=-500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex]
GATE EE 2015-SET-1   Power Electronics
Question 3 Explanation: 
To maintain the direction of power flow from system 2 to system 1, the voltage V_1=-485kV and voltage V_2=500kV and I=1.5kA.
Since, current cannot flow in reverse direction. Option (B) is correct answer.
Question 4
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 \mus is applied to the SCR. The maximum value of R in \Omega to ensure successful firing of the SCR is ______.
A
4050
B
5560
C
6060
D
8015
GATE EE 2014-SET-2   Power Electronics
Question 4 Explanation: 
Let us assume the SCR is conducting,

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

V=I \times R
R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega
Question 5
A single-phase SCR based ac regulator is feeding power to a load consisting of 5 \Omega resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are
A
30^{\circ} and 46 A
B
30^{\circ} and 23 A
C
45^{\circ} and 23 A
D
45^{\circ} and 32 A
GATE EE 2014-SET-2   Power Electronics
Question 5 Explanation: 
The maximum firing angle at which the voltage across the device becomes '\phi ' = load angle.
\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}
Rms value of current through SCR is
\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}
Question 6
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10\Omega.

The kVA rating of the input transformer is
A
53.2 kVA
B
46.0 kVA
C
22.6 kVA
D
7.5 kVA
GATE EE 2011   Power Electronics
Question 6 Explanation: 
RMS value of supply current in case of 3-\phi bridge converter
I_s=I_0\sqrt{\frac{2}{3}}=40\sqrt{\frac{2}{3}}=32.66A
KVA rating of the input transformrer
\begin{aligned} &=\sqrt{3}V_sI_s \\ &= \sqrt{3} \times 400 \times 32.66 \times 10^{-3}\; kVA\\ &=22.62\; kVA \end{aligned}
Question 7
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10\Omega.

The maximum current through the battery will be
A
14A
B
40A
C
80A
D
94A
GATE EE 2011   Power Electronics
Question 7 Explanation: 


Average output voltage of the converter
V_0=\frac{3 V_{ml}}{\pi} \cos \alpha
The converter acts as line commutated inverter and for such mode \alpha \gt 90^{\circ} and V_0 is negative.THerefore, battery supplies energy to AC system. So, current through battery
I_0=\frac{400-V_0}{R}
For V_0=0 \text{ or }\alpha =90^{\circ},
Maximum current flow through battery
(I_0)_{max}=\frac{400}{10}=40A
Question 8
The input voltage given to a converter is
v_i=100\sqrt{2}\sin (100 \pi t) V
The current drawn by the converter is
i_i=10\sqrt{2}\sin (100\pi t-\pi/3) + 5\sqrt{2}\sin (300\pi t+\pi/4) + 2\sqrt{2}\sin (500\pi t-\pi/6)A
The active power drawn by the converter is
A
181W
B
500W
C
707W
D
887W
GATE EE 2011   Power Electronics
Question 8 Explanation: 
Rms value of input voltag,
V_{rms}=\frac{100\sqrt{2}}{\sqrt{2}}=100V
Rms value of current,
I_{rms}=\sqrt{\left (\frac{10\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{5\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{2\sqrt{2}}{\sqrt{2}} \right )^2}=11.358A
Let input power factor \cos \phi
V_{rms}I_{rms} \cos \phi = active power drawn by the coverter
\Rightarrow \;100 \times 11.358 \times \cos \phi =500W
\Rightarrow \; \cos \phi =0.44
Question 9
The input voltage given to a converter is
v_i=100\sqrt{2}\sin (100 \pi t) V
The current drawn by the converter is
i_i=10\sqrt{2}\sin (100\pi t-\pi/3) + 5\sqrt{2}\sin (300\pi t+\pi/4) + 2\sqrt{2}\sin (500\pi t-\pi/6)A
The input power factor of the converter is
A
0.31
B
0.44
C
0.5
D
0.71
GATE EE 2011   Power Electronics
Question 9 Explanation: 
\begin{aligned} V_i&=100\sqrt{2}\sin (100 \pi t)\\ i_i&=10\sqrt{2} \sin \left ( 100 \pi t -\frac{\pi}{3} \right )\\ &+5\sqrt{2} \sin \left ( 300 \pi t +\frac{\pi}{4} \right )\\ &+2\sqrt{2} \sin \left ( 500 \pi t -\frac{\pi}{4} \right )A \end{aligned}
Fundamental component of input voltage
\begin{aligned} (V_i)_1&=100\sqrt{2} \sin (100 \pi t)\\ (V_i)_{1,rms}&=\frac{100\sqrt{2}}{\sqrt{2}}=100V \end{aligned}
Fundamental component of current
\begin{aligned} (i_L)_1&=10\sqrt{2} \sin (100 \pi t-\frac{\pi}{3})\\ (i_L)_{1,rms}&=\frac{10\sqrt{2}}{\sqrt{2}}=10 \end{aligned}
Phase difference between these two components
\phi _1=\frac{\pi}{3}, \cos \phi _1= \cos \frac{\pi}{3}=0.5
Active power due to fundamental components
P_1=(V_i)_{1,rms} \times (i_i)_{1,rms} \cos \phi =100 \times 10 \times 0.5=500W
Since 3^{rd} \text{ and } 5^{th} harmonic are absent in input voltage, there is no active power due to the these components.
Hence, active power drawn by the converter
P_0= Active power due to fundamental components =500 W
Question 10
The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a
A
step down chopper (buck converter)
B
half-wave rectifier
C
step-up chopper (boost converter)
D
full-wave rectifier
GATE EE 2010   Power Electronics
Question 10 Explanation: 
When switch is connected to A for time duration T_1

V_{out}=V_{in}
When switch is connected to B for time duration T_2

Average output voltage =\frac{V_{in}T_1}{T_1+T_2}=\alpha V_{in}
where, \alpha =\text{duty cycle}=\frac{T_1}{T_1+T_2}
Therefore, the converter shown is a step down chooper.
There are 10 questions to complete.

GATE Electrical Engineering-Topic wise Previous Year Questions

Prepare for GATE 2023 with practice of GATE Electrical previous year questions and solution

Prepare for GATE 2023 with practice of GATE Electrical previous year questions and solution

GATE 2022 Electrical Engineering Syllabus

Revised syllabus of GATE 2022 Electrical Engineering by IIT.

Practice GATE Electrical Engineering previous year questions

Year wise | Subject wise | Topic wise

Section 1: Engineering Mathematics

Linear Algebra: Matrix Algebra, Systems of linear equations, Eigenvalues, Eigenvectors.
Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series, Vector identities, Directional derivatives, Line integral, Surface integral, Volume integral, Stokes’s theorem, Gauss’s theorem, Divergence theorem, Green’s theorem.
Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s equation, Euler’s equation, Initial and boundary value problems, Partial Differential Equations, Method of separation of variables.
Complex variables: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula, Taylor series, Laurent series, Residue theorem, Solution integrals.
Probability and Statistics: Sampling theorems, Conditional probability, Mean, Median, Mode, Standard Deviation, Random variables, Discrete and Continuous distributions, Poisson distribution, Normal distribution, Binomial distribution, Correlation analysis, Regression analysis.
Section 2: Electric circuits

Network elements: ideal voltage and current sources, dependent sources, R, L, C, M elements; Network solution methods: KCL, KVL, Node and Mesh analysis; Network Theorems: Thevenin’s, Norton’s, Superposition and Maximum Power Transfer theorem; Transient response of dc and ac networks, sinusoidal steady-state analysis, resonance, two port networks, balanced three phase circuits, star-delta transformation, complex power and power factor in ac circuits.
Section 3: Electromagnetic Fields

Coulomb’s Law, Electric Field Intensity, Electric Flux Density, Gauss’s Law, Divergence, Electric field and potential due to point, line, plane and spherical charge distributions, Effect of dielectric medium, Capacitance of simple configurations, Biot‐Savart’s law, Ampere’s law, Curl, Faraday’s law, Lorentz force, Inductance, Magnetomotive force, Reluctance, Magnetic circuits, Self and Mutual inductance of simple configurations.
Section 4: Signals and Systems

Representation of continuous and discrete time signals, shifting and scaling properties, linear time invariant and
causal systems, Fourier series representation of continuous and discrete time periodic signals, sampling theorem,
Applications of Fourier Transform for continuous and discrete time signals, Laplace Transform and Z transform.
Section 5: Electrical Machines

Single phase transformer: equivalent circuit, phasor diagram, open circuit and short circuit tests, regulation and
efficiency;
Three-phase transformers: connections, vector groups, parallel operation; Auto-transformer, Electromechanical energy conversion principles;
DC machines: separately excited, series and shunt, motoring and generating mode of operation and their characteristics, speed control of dc motors;
Three-phase induction machines: principle of operation, types, performance, torque-speed characteristics, no-load and blocked-rotor tests, equivalent circuit, starting and speed control; Operating principle of single-phase induction motors;
Synchronous machines: cylindrical and salient pole machines, performance and characteristics, regulation and parallel operation of generators, starting of synchronous motors; Types of losses and efficiency calculations of electric machines
Section 6: Power Systems

Basic concepts of electrical power generation, ac and dc transmission concepts, Models and performance of transmission lines and cables, Series and shunt compensation, Electric field distribution and insulators, Distribution systems, Per‐unit quantities, Bus admittance matrix, Gauss- Seidel and Newton-Raphson load flow methods, Voltage and Frequency control, Power factor correction, Symmetrical components, Symmetrical and unsymmetrical fault analysis, Principles of over‐current, differential, directional and distance protection; Circuit breakers, System stability concepts, Equal area criterion, Economic Load Dispatch (with and without considering transmission losses).
Section 7: Control Systems

Mathematical modeling and representation of systems, Feedback principle, transfer function, Block diagrams and Signal flow graphs, Transient and Steady‐state analysis of linear time invariant systems, Stability analysis using Routh-Hurwitz and Nyquist criteria, Bode plots, Root loci, Lag, Lead and Lead‐Lag compensators; P, PI and PID controllers; State space model, Solution of state equations of LTI systems, R.M.S. value, average value calculation for any general periodic waveform.
Section 8: Electrical and Electronic Measurements

Bridges and Potentiometers, Measurement of voltage, current, power, energy and power factor; Instrument transformers, Digital voltmeters and multimeters, Phase, Time and Frequency measurement; Oscilloscopes, Error analysis.
Section 9: Analog Electronics and Digital Electronics

Simple diode circuits: clipping, clamping, rectifiers; Amplifiers: biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers: characteristics and applications; single stage active filters, Sallen Key, Butterworth, VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers, Schmitt triggers, sample and hold circuits, A/D and D/A converters.
Section 10: Power Electronics

Static V-I characteristics and firing/gating circuits for Thyristor, MOSFET, IGBT; DC to DC conversion: Buck, Boost and Buck-Boost Converters; Single and three-phase configuration of uncontrolled rectifiers; Voltage and Current commutated Thyristor based converters; Bidirectional ac to dc voltage source converters; Magnitude and Phase of line current harmonics for uncontrolled and thyristor based converters; Power factor and Distortion Factor of ac to dc converters; Single-phase and three-phase voltage and current source inverters, sinusoidal pulse width modulation.

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GATE EE 2013

Question 1
In the circuit shown below what is the output voltage (V_{out}) if a silicon transistor Q and an ideal op-amp are used?
A
-15 V
B
-0.7 V
C
+0.7 V
D
+15 V
Analog Electronics   Operational Amplifiers
Question 1 Explanation: 


Using the concept of vitual ground, V=0

V_{out}=-0.7V
Question 2
The transfer function \frac{V_{2}(s)}{V_{1}(s)} of the circuit shown below is
A
\frac{0.5s+1}{s+1}
B
\frac{3s+6}{s+2}
C
\frac{s+2}{s+1}
D
\frac{s+1}{s+2}
Control Systems   Mathematical Models of Physical Systems
Question 2 Explanation: 


\frac{V_2(s)}{V_1(s)}=\frac{R+\frac{1}{Cs}}{\frac{1}{Cs}+R+\frac{1}{Cs}}
=\frac{1+RCs}{2+RCs}
=\frac{1+10 \times 10^3 \times 100 \times 10^{-6}s}{2+10 \times 10^3 \times 100 \times 10^{-6}s}
=\frac{s+1}{s+2}
Question 3
Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is
A
u(t)
B
tu(t)
C
\frac{t^{2}}{2}u(t)
D
e^{-t}u(t)
Signals and Systems   Laplace Transform
Question 3 Explanation: 
\begin{aligned} Y(s)&=\frac{1}{s}U(s)=\frac{1}{s^2}\\ y(t)&=tu(t) \end{aligned}
Question 4
The impulse response of a system is h(t)=tu(t). For an input u(t-1), the output is
A
\frac{t^{2}}{2}u(t)
B
\frac{t(t-1)}{2}u(t-1)
C
\frac{(t-1)^{2}}{2}u(t-1)
D
\frac{t^{2}-1}{2}u(t-1)
Signals and Systems   Linear Time Invariant Systems
Question 4 Explanation: 
\begin{aligned} h(t) &=tu(t) \\ H(s) &=\frac{1}{s^2} \\ \Rightarrow \; \frac{Y(s)}{U(s)} &=\frac{1}{s^2} \\ Y(s) &=\frac{1}{s^2} \frac{e^{-s}}{s}\\ \Rightarrow \; y(t) &=\frac{(t-1)^2}{2}u(t-1) \end{aligned}
Question 5
Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system?
A
All the poles of the system must lie on the left side of the j \omega axis
B
Zeros of the system can lie anywhere in the s-plane
C
All the poles must lie within |s| = 1
D
All the roots of the characteristic equation must be located on the left side of the j \omega axis.
Signals and Systems   Laplace Transform
Question 5 Explanation: 
All poles must lie within |Z|=1
Question 6
Two systems with impulse responses h_{1}(t) \; and \; h_{2}(t) are connected in cascade. Then the overall impulse response of the cascaded system is given by
A
product of h_{1}(t) \; and \; h_{2}(t)
B
sum of h_{1}(t) \; and \; h_{2}(t)
C
convolution of h_{1}(t) \; and \; h_{2}(t)
D
subtraction of h_{1}(t) \; and \; h_{2}(t)
Signals and Systems   Linear Time Invariant Systems
Question 6 Explanation: 
In cascade connection,

\begin{aligned} H(s) &=H_1(s)\cdot H_2(s)\\ \Rightarrow \; h(t)&=h_1(t) * h_2(t) \end{aligned}
Question 7
A source v_{s}(t)=V \cos 100\pi t has an internal impedance of (4+j3)\Omega. If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in \Omega should be
A
3
B
4
C
5
D
7
Electric Circuits   Network Theorems
Question 7 Explanation: 
Using maximum power transfer theorem,
R_L=|Z|=|4-j3|
\;\;=\sqrt{4^2+3^2}=5\Omega
Question 8
A single-phase load is supplied by a single-phase voltage source. If the current flowing from the load to the source is 10\angle -150^{\circ}A and if the voltage at the load terminal is 100\angle 60^{\circ}V, then the
A
load absorbs real power and delivers reactive power
B
load absorbs real power and absorbs reactive power
C
load delivers real power and delivers reactive power
D
load delivers real power and absorbs reactive power
Power Systems   Load Flow Studies
Question 8 Explanation: 


Complex power supplied by load =(100\angle 60^{\circ})(10\angle -150^{\circ})=-866.022-j500 VA
As supplied active and reactive power are negative.
Load absorbs active and reactive power both.
Question 9
A single-phase transformer has no-load loss of 64W, as obtained from an open circuit test. When a short-circuit test is performed on it with 90% of the rated currents flowing in its both LV and HV windings, he measured loss is 81 W. The transformer has maximum efficiency when operated at
A
50.0% of the rated current
B
64.0% of the rated current
C
80.0% of the rated current
D
88.8% of the rated current
Electrical Machines   Transformers
Question 9 Explanation: 
For 90% current,
\begin{aligned} P_{cu} &=(0.9)^2 P_{fl\;cu} \\ P_{fl\;cu} &= \frac{81}{0.81}=100W\\ x&=\sqrt{\frac{P_{core}}{P_{fl\;cu}}}=\sqrt{\frac{64}{100}}=0.8=80\% \end{aligned}
Question 10
The flux density at a point in space is given by B=4xa_{x}+2kya_{y}+8a_{z} Wb/m^{2}. The value of constant k must be equal to
A
-2
B
-0.5
C
0.5
D
2
Electromagnetic Fields   Magnetostatic Fields
Question 10 Explanation: 
\begin{aligned} \bigtriangledown \cdot B &=0 \\ \left ( \frac{\partial }{\partial x}a_x +\frac{\partial }{\partial y}a_y + \frac{\partial }{\partial z}a_z\right )&(4xa_x+2kya_y+8a_z)=0 \\ 4+2k&=0\\ k&=-2 \end{aligned}
There are 10 questions to complete.

GATE EE 2014 SET 1

Question 1
Given a system of equations
x + 2y + 2z = b_1
5x + y + 3z = b_2
What of the following is true regarding its solutions
A
The system has a unique solution for any given b_1 \; and \; b_2
B
The system will have infinitely many solutions for any given b_1 \; and \; b_2
C
Whether or not a solution exists depends on the given b_1 \; and \; b_2
D
The systems would have no solution for any values of b_1 \; and \; b_2
Engineering Mathematics   Linear Algebra
Question 2
Let f(x)=xe^{-x}. The maximum value of the function in the interval (0,\infty ) is
A
e^{-1}
B
e
C
1-e^{-1}
D
1+e^{-1}
Engineering Mathematics   Calculus
Question 2 Explanation: 
\begin{aligned} f(x)&=xe^{-x}\\ f'(x)&=e^{-x}-xe^{-x}=0\\ e^{-x}(1-x)&=0\\ x&=1\\ f''(x)&=-e^{-x}-e^{-x}+xe^{-x}\\ &=e^{-x}(x-2)\\ f''(1)&=e^{-1}(-1)=-e^{-1} \lt 0 \end{aligned}
Hence f(x) has maximum value at x=1
f(1)=1\cdot e^{-1}=e^{-1}
Question 3
The solution for the differential equation
\frac{d^{2}x}{dt^{2}}=-9x
with initial conditions x(0)=1 and \frac{dx}{dt}|_{t=0}=1, is
A
t^{2}+t+1
B
sin 3t+ \frac{1}{3} cos3t+\frac{2}{3}
C
\frac{1}{3} sin 3t+cos3t
D
cos3t+t
Engineering Mathematics   Differential Equations
Question 3 Explanation: 
\begin{aligned} \frac{d^2x}{dt^2}&=-9x\\ \frac{d^2x}{dt^2}+9x&=0\\ (D^2+9)x &=0 \end{aligned}
Auxiliary equation is m^2+9=0
\begin{aligned} m&= \pm 3i\\ x&= C_1 \cos 3t+C_2 \sin 3t\;\;...(i)\\ x(0) &=1\;\;i.e.\;x\rightarrow 1\; when t\rightarrow 0 \\ 1&=C_1 \\ \frac{dx}{dt}&=-3C_1 \sin 3t+3C_2 \cos 3t\;\;...(ii) \\ x'(0)&=1\;\; i.e.\;x'\rightarrow 1\; when \; t\rightarrow 0 \\ 1&=3C_2 \\ C_2&=\frac{1}{3}\\ \therefore \;x&=\cos 3t+\frac{1}{3}\sin 3t \end{aligned}
Question 4
Let X(s)=\frac{3s+5}{s^{2}+10s+21} be the Laplace Transform of a signal x(t). Then, x(0^+) is
A
0
B
3
C
5
D
21
Signals and Systems   Laplace Transform
Question 4 Explanation: 
Given, X(s)=\left [ \frac{3s+5}{s^2+10s+21} \right ]
Using initial value theorem,
\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}
Question 5
Let S be the set of points in the complex plane corresponding to the unit circle. (That is, S={z:|z|=1}). Consider the function f(z)=zz* where z* denotes the complex conjugate of z. The f(z) maps S to which one of the following in the complex plane
A
unit circle
B
horizontal axis line segment from origin to (1, 0)
C
the point (1, 0)
D
the entire horizontal axis
Engineering Mathematics   Complex Variables
Question 5 Explanation: 


\begin{aligned} z&=x+iy\\ z^*&=x-iy\\ zz^*&=(x+iy)(x-iy)\\ &=x^2+y^2 \end{aligned}
which is equal to (1) always as given
\begin{aligned} |z|&=1\\ zz^*&=x^2+y^2 \end{aligned}

Question 6
The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is _____.
A
110
B
250
C
330
D
360
Electric Circuits   Basics
Question 6 Explanation: 
Given electrical circuit is shown below:

Applying KCL at node, current through 15V voltage source =2 A
Power absorbed by 100 V voltage source = 10x100=1000 Watt.
Power absorbed by 80 V voltage source =-(80x8)=-640 Watts and power absorbed by 15 V voltage source = -(15x2)=-30 Watt.
Therefore, total power absorbed by the three circuit element=(100-640-30_ watts =330 Watts
Question 7
C_0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity \epsilon _r , the expression for the modified capacitance is
A
\frac{C_{0}}{2}(1+\epsilon _{r})
B
(C_{0}+\epsilon _{r})
C
\frac{C_{0}}{2}(\epsilon _{r})
D
C_{0}(1+\epsilon _{r})
Electromagnetic Fields   Electrostatic Fields
Question 7 Explanation: 
Let A be the area of the parallel plate capacitor and d be the distance between the plates.

Let C_1 be the capacitance of half portion with air as dielectic medium and C_2 be capacitance with a dielectic of permitivity, \varepsilon _r.
Then, C_1=\frac{\varepsilon _0 \left ( \frac{A}{2} \right )}{d} (As area become half)
and C_2=\frac{\varepsilon _0 \varepsilon _r \left ( \frac{A}{2} \right )}{d} (As area becomes half)
Now, these two capacitance will be in parallel if a voltage is applied between the plates as same potential difference will be there between both the capacitance.

Equivalent capacitance is
\begin{aligned} C_{eqv} &= C_1+C_2\\ &=\frac{\varepsilon _0 A}{2d}+ \frac{\varepsilon _0 \varepsilon _r A}{2d}\\ &= \frac{\varepsilon _0 A}{2d}(1+ \varepsilon _r)=\frac{C_0}{2}(1+\varepsilon _r) \end{aligned}
Therefore, modified capacitance,
C_{eqv}=\frac{C_0}{2}(1+\varepsilon _r)
Question 8
A combination of 1 \muF capacitor with an initial voltage v_c(0)=-2V in series with a 100 \Omega resistor is connected to a 20 mA ideal dc current source by operating both switches at t=0s as shown. Which of the following graphs shown in the options approximates the voltage v_s across the current source over the next few seconds ?

A
A
B
B
C
C
D
D
Electric Circuits   Transients and Steady State Response
Question 8 Explanation: 
Given C=1\mu F, V_c(0)=-2V, R=100\Omega , I=20mA. Circuit fot the given condition at time t \gt 0 is shown below:

Applying KVL, we have,V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )
\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]
\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]
Putting values of R, C and I, we get,
V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]
\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]
\;\;=\frac{20 \times 10^3}{s^2}
\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}
V_s(t)=20000 t u(t)
\therefore \;\; V_s(s)=(20000)tu(t)
Which is equation of a straight line passing through origin. Hence option (C) is correct.
Question 9
x(t) is nonzero only for T_x \lt t \lt T'_x, and similarly, y(t) is nonzero only for T_y \lt t \lt T'_y. Let z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE ?
A
z(t) can be nonzero over an unbounded interval.
B
z(t) is nonzero for t \lt T_{x}+T_{y}
C
z(t) is zero outside of T_{x}+T_{y} \lt t \lt T'_{x}+T'_{y}
D
z(t) is nonzero for t\gt T'_{x}+T'_{y}
Signals and Systems   Linear Time Invariant Systems
Question 9 Explanation: 
x(t) is non-zero for T_x \lt t \lt T'_x
and y(t) is non-zero for T_y \lt t \lt T'_y
\because \;\; z(t)=x(t)\otimes y(t)
then the limits of the resultant signal is
T_x+T_y \lt t \lt T'_x+T'_y
i.e. z(t) is non-zero for T_x+T_y \lt t \lt T'_x+T'_y
Question 10
For a periodic square wave, which one of the following statements is TRUE ?
A
The Fourier series coefficients do not exist
B
The Fourier series coefficients exist but the reconstruction converges at no point
C
The Fourier series coefficients exist but the reconstruction converges at most point
D
The Fourier series coefficients exist and the reconstruction converges at every point.
Signals and Systems   Fourier Series
Question 10 Explanation: 


Reconstruction of signal by its Fourier series coefficient is not possible at those points where signal is discontinuous.
In the above figure, at integer multiples of 'T/2', signal recovery is not possible by using its coefficient.
Therefore, reconstruction of x(t) by using its coefficient is possible at most of the points except those instants where x(t) is discontinous.
There are 10 questions to complete.

GATE EE 2014 SET 2

Question 1
Which one of the following statements is true for all real symmetric matrices?
A
All the eigenvalues are real.
B
All the eigenvalues are positive.
C
All the eigenvalues are distinct.
D
Sum of all the eigenvalues is zero.
Engineering Mathematics   Linear Algebra
Question 2
Consider a dice with the property that the probability of a face with n dots showing up proportional to n. The probability of the face with three dots showing up is____.
A
0.1
B
0.33
C
0.14
D
0.66
Engineering Mathematics   Probability and Statistics
Question 2 Explanation: 
Let probability of occurence of one dot is P.
So, writing total probability
P+2P+3P+4P+5P+6P=1
P=\frac{1}{21}
Hence, problem of occurrence of 3 dot is =3P=\frac{3}{21}=\frac{1}{7}=0.142
Question 3
Minimum of the real valued function f(x)=(x-1)^{2/3} occurs at x equal to
A
-\infty
B
0
C
1
D
\infty
Engineering Mathematics   Calculus
Question 3 Explanation: 
f(x)=(x-1)^{2/3}=(\sqrt[3]{x-1})^2
As f(x) is square of \sqrt[3]{x-1}, hence its minimum value be 0 where at x=1.
Question 4
All the values of the multi-valued complex function 1^i, where i=\sqrt{-1}, are
A
purely imaginary
B
real and non-negative
C
on the unit circle
D
equal in real and imaginary parts
Engineering Mathematics   Complex Variables
Question 4 Explanation: 
Let z=1^i=1^{e^{i(4n+1)\pi/2}}\;\;\;n \in I
z=1 which is purly real and non negative.
Question 5
Consider the differential equation x^{2}\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}-y=0. Which of the following is a solution to this differential equation for x \gt 0 ?
A
e^{x}
B
x^{2}
C
1/x
D
ln x
Engineering Mathematics   Differential Equations
Question 5 Explanation: 
\begin{aligned} x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y&=0\\ \text{Let, }x=e^z\leftrightarrow z&=\log x\\ s\frac{d}{dx}=xD=\theta &=\frac{d}{dz}\\ x^2D^2&=\theta (\theta -1)\\ (x^2D^2+xD-1)y&=0\\ [\theta (\theta -1)+\theta -1]y&=0\\ (\theta ^2-\theta +\theta -1)&=0\\ (\theta ^2-1)y&=0\\ \text{Auxiliary equation is }m^2-1&=0\\ m&=\pm 1\\ \text{CF is }C_1e^{-z}+C_2e^z&\\ \text{Solution is }y&=C_1e^{-z}+C_2e^z\\ y&=C_1x^{-1}+C_2x\\ y&=C_1\frac{1}{x}+C_2x \end{aligned}
One independent solution is \frac{1}{x}
Another independent solution is x.
Question 6
Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 \muH and 240 \muH. Their mutual inductance in \muH is _____.
A
10
B
15
C
55
D
35
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 6 Explanation: 
The two possible series connection are shown below:
Let the mutual inductance be M

(i) Additive connection,

(ii) Substractive connection,
L_{eqv}=L_1+L_2-2M=240\mu H
Thus, L_1+L_2+2M=380 \mu H\;\;...(i)
and L_1+L_2-2M=240 \mu H\;\;...(ii)
Solving equations (i) and (ii), we get:
4M=140\mu H
M=35\mu H
Therefore, mutual inductance M=35\mu H
Question 7
The switch SW shown in the circuit is kept at position '1' for a long duration. At t=0+, the switch is moved to position '2'. Assuming |V_{o2} | \gt |V_{o1}|, the voltage v_{c}(t) across the capacitor is
A
v_{c}(t)=-V_{o2}(1-e^{-t/2RC})-V_{o1}
B
v_{c}(t)=V_{o2}(1-e^{-t/2RC})+V_{o1}
C
v_{c}(t)=-(V_{o2}+V_{o1})(1-e^{-t/2RC})-V_{o1}
D
v_{c}(t)=(V_{o2}-V_{o1})(1-e^{-t/2RC})+V_{o1}
Electric Circuits   Transients and Steady State Response
Question 8
A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates.

If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio \varepsilon _1:\varepsilon _2 is
A
1:4
B
2:3
C
3:2
D
4:1
Electromagnetic Fields   Electrostatic Fields
Question 8 Explanation: 
Let, A = Area of plates
Let C_1= C_3 be the capacitance formed with dielectric having dielectric constant \varepsilon _1.
C_{eqv} be the equivalent capacitance.
C_2 be the capacitance formed with dielectic having dielectric constant \varepsilon _{r_2}.
Then, C_1=C_3=\frac{\varepsilon _0\varepsilon _1A}{\frac{d}{4}}=\frac{4\varepsilon _0\varepsilon _1A}{d}
and C_2=\frac{\varepsilon _0\varepsilon _2A}{\frac{d}{2}}=\frac{2\varepsilon _0\varepsilon _2A}{d}
Also, equivalent capacitance = C_{eqv}
\begin{aligned} \therefore \;\;\frac{1}{C_{eqv}}&=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{2}{C_1}+\frac{1}{C_2} \\ &(\because \;\;C_1=C_3)(C_1,C_2,C_3 \;\text{all in series}) \\ \frac{1}{C_{eqv}} &= \frac{2d}{4\varepsilon _0\varepsilon _1d}+\frac{d}{2\varepsilon _0\varepsilon _2d}\\ C_{eqv}&=\frac{2\varepsilon _1\varepsilon _2\varepsilon _0A}{d(\varepsilon _1+\varepsilon _2)} \end{aligned}
Given, V_{eqv}= Total Voltage = 10 VOlt,
V_1=V_3=2 Volt
We know that : C\propto \frac{1}{V}
\begin{aligned} &\therefore \; \frac{C_{eqv}}{C_1}=\frac{V_1}{V_{eqv}}=\frac{1}{5} \\ &\frac{2\varepsilon _0\varepsilon _1\varepsilon _2 A}{d(\varepsilon _1+\varepsilon _2)}\times \frac{d}{4\varepsilon _0\varepsilon _1A}=\frac{1}{5} \\ &\frac{\varepsilon _2}{2(\varepsilon _1+\varepsilon _2)}=\frac{1}{5} \\ &5\varepsilon _2=2\varepsilon _1+2\varepsilon _2 \\ &2\varepsilon _1 =3\varepsilon _2\\ &\varepsilon _1:\varepsilon _2=3:2 \end{aligned}
Question 9
Consider an LTI system with transfer function
H(s)=\frac{1}{s(s+4)}
If the input to the system is cos(3t) and the steady state output is Asin(3t+\alpha) , then the value of A is
A
1/30
B
1/15
C
3/4
D
4/3
Signals and Systems   Laplace Transform
Question 9 Explanation: 
Given,
\begin{aligned} H(s)&=\frac{1}{s(s+4)}\\ r(t)&=\text{input}= \cos (3t)=\cos \omega t\\ \therefore \; \omega &=3 \text{ rad/s}\\ H(j\omega)&=\frac{1}{(j\omega)(j\omega+4)}\\ \text{Now, }|H(j\omega)|&=\frac{1}{\omega \sqrt{\omega^4+4^2}}\\ &=\frac{1}{3\sqrt{25}}=\frac{1}{15} \;\;\;(at\; \omega=3)\\ \angle H(j\omega)&=-90^{\circ}-\tan ^{-1}\frac{\omega}{4}\\ &=-90^{\circ}-\tan ^{-1}\frac{3}{4}=-126.86^{\circ}\\ \therefore \; c(t)&=\frac{1}{15} \cos (3t-126.86^{\circ})\\ &=\frac{1}{15} \sin (3t-36.86^{\circ})\;\;...(i)\\ c(t)&=A \sin (3t+\alpha )\;\;...(ii)\\ \text{Comparing }&\text{ eq. (i) and (ii), we have,}\\ A&=\frac{1}{15} \end{aligned}
Question 10
Consider an LTI system with impulse response h(t)=e^{-5t}u(t). If the output of the system is y(t)=e^{-3t}u(t)-e^{-5t}u(t) then the input, x(t), is given by
A
e^{-3t}u(t)
B
2e^{-3t}u(t)
C
e^{-5t}u(t)
D
2e^{-5t}u(t)
Signals and Systems   Linear Time Invariant Systems
Question 10 Explanation: 
Impulse response of an LTI system = transfer function =\frac{Y(s)}{X(s)}=H(s)
\begin{aligned} \text{where, } y(t)&=e^{-3t}u(t)-e^{-5t}u(t) \\ \therefore \; Y(s) &=\frac{1}{(s+3)}-\frac{1}{(s+5)} \\ &=\frac{2}{(s+3)(s+5)} \\ \text{Also, } h(t)&=e^{-5t}u(t) \\ \therefore \; H(s) &=\frac{1}{(s+5)} \\ \text{Therefore, } X(s)&=\frac{Y(s)}{H(s)} \\ &=\frac{2}{(s+3)(s+5)} \times (s+5) \\ &= \frac{2}{(s+3)}\\ \therefore \; \text{Input}&=x(t)=2e^{-3t}u(t) \end{aligned}
There are 10 questions to complete.

GATE EE 2014 SET 3

Question 1
Two matrices A and B are given below:
A=\begin{bmatrix} p &q \\ r& s \end{bmatrix}; B=\begin{bmatrix} p^2+q^2 & pr +qs \\ pr+qs & r^2+s^2 \end{bmatrix}
If the rank of matrix A is N, then the rank of matrix B is
A
N/2
B
N-1
C
N
D
2N
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\begin{aligned} A &=\begin{bmatrix} p & q\\ r & s \end{bmatrix} \\ A \times A=A^2&=\begin{bmatrix} p^2+q^2 & pr+qs\\ pr+qs & r^2+s^2 \end{bmatrix} =B \\ A^2 &=B \end{aligned}
Rank of amtrix does not change when we squaring the matrix, hence rank of B = rank of A=N.
Question 2
A particle, starting from origin at t=0s, is traveling along x-axis with velocity
v=\frac{\pi}{2}\cos (\frac{\pi}{2}t)m/s
At t=3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_____
A
1
B
2
C
3
D
4
Engineering Mathematics   Calculus
Question 3
Let \triangledown \cdot (f v)=x^2y+y^2z+z^2x, where f and v are scalar and vector fields respectively. If v=yi+zj+xk , then v\cdot \triangledown f is
A
x^2y+y^2z+z^2x
B
2xy+2yz+2zx
C
x+y+z
D
0
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} \vec{V}&=y\hat{i}+z\hat{j}+x\hat{k}\\ \hat{i}\frac{\partial (fV)}{\partial x}+\hat{j}\frac{\partial (fV)}{\partial y}+\hat{k}\frac{\partial (fV)}{\partial z}&=x^2y+y^2z+z^2x\\ y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}&=x^2y+y^2z+z^2x\;\;...(i)\\ \vec{V}\cdot \Delta f&=y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}\;\;...(ii)\\ \text{From equations (i) and (ii)}\\ \vec{V}\cdot \Delta f&=x^2y+y^2z+z^2x \end{aligned}
Question 4
Lifetime of an electric bulb is a random variable with density f(x)=kx^2, where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is _____
A
0.85
B
0.42
C
0.25
D
0.75
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
Life time of an electric bulb with density
f(x)=Kx^2
If minimum and maximum lifetimes of bulb are 1 and 2 years respectively then
\begin{aligned} \int_{1}^{2}Kx^2dx &=1\\ \left.\begin{matrix} K\frac{x^3}{3} \end{matrix}\right|_1^2&=1\\ K\left ( \frac{8}{3}-\frac{1}{3} \right )&=1\\ \frac{7K}{3}&=1\\ K&=\frac{3}{7}=0.42 \end{aligned}
Question 5
A function f(t) is shown in the figure.

The Fourier transform F(\omega) of f(t) is
A
real and even function of w
B
real and odd function of w
C
imaginary and odd function of w
D
imaginary and even function of w
Signals and Systems   Fourier Transform
Question 5 Explanation: 
Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .
Question 6
The line A to neutral voltage is 10 \angle 15^{\circ}V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by
A
10 \sqrt{3}\angle 105^{\circ} V
B
10 \angle 105^{\circ} V
C
10 \sqrt{3}\angle -75^{\circ} V
D
-10 \sqrt{3}\angle 90^{\circ} V
Electric Circuits   Three-Phase Circuits
Question 6 Explanation: 
Given,
V_{AN}=10\angle 15^{\circ}volt
As the system is balanced and phase sequence is ABC, therefore,
V_{AN}=10\angle 15^{\circ}
V_{BN}=10\angle 135^{\circ}V and V_{CN}=10\angle 255^{\circ}V
\therefore voltage of line w.r.t. line C is
V_{BC}=V_{BN}-V_{CN}
V_{BC}=10\angle 255^{\circ}-10\angle 135^{\circ}
V_{BC}=10\sqrt{3}\angle -75^{\circ} Volt
Question 7
A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R(\gt r) is
A
4\pi \varepsilon _{0}r
B
4\pi \varepsilon _{0}r^{2}
C
4\pi \varepsilon _{0}R
D
4\pi \varepsilon _{0}R^{2}
Electromagnetic Fields   Electrostatic Fields
Question 8
The driving point impedance Z(s) for the circuit shown below is
A
\frac{s^{4}+3s^{2}+1}{s^{3}+2s}
B
\frac{s^{4}+3s^{2}+4}{s^{2}+2}
C
\frac{s^{2}+1}{s^{4}+s^{2}+1}
D
\frac{s^{3}+1}{s^{4}+s^{2}+1}
Electric Circuits   Two Port Network and Network Functions
Question 8 Explanation: 


Driving point impedance, Z(s) is ,
Z(s)=s+\left ( \frac{\left ( s+\frac{1}{s} \right )\times \frac{1}{s}}{s+\frac{1}{s}+\frac{1}{s}} \right )
\;\;=s+\left ( \frac{s^2+1}{s^2} \right )\times \frac{s}{s^2+2}
\;\;=s+\frac{s^2+1}{s(s^2+2)}
\;\;=\frac{s^2(s^2+2)+^2+1}{s^3+2s}
Z(s)=\frac{s^4+3s^2+1}{s^3+2s}
Question 9
A signal is represented by
x(t)=\left\{\begin{matrix} 1 &|t| \lt 1 \\ 0& |t|\gt 1 \end{matrix}\right.
The Fourier transform of the convolved signal y(t)= x(2t)* x(t/2) is
A
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})sin(2 \omega )
B
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})
C
\frac{4}{\omega ^{2}}sin(2 \omega )
D
\frac{4}{\omega ^{2}}sin^{2} \omega
Signals and Systems   Fourier Transform
Question 9 Explanation: 
Given signal can be drawn as

Therefore,
\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}
Convolution in time domain multiplication in frequency domain
\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}
Question 10
For the signal
f(t) = 3 \sin 8 \pi t + 6 \sin 12 \pi t + \sin 14 \pi t,
the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____.
A
7
B
14
C
18
D
9
Signals and Systems   Sampling
Question 10 Explanation: 
\begin{aligned} f_{m_1} &=4Hz \\ f_{m_2} &=6Hz \\ f_{m_3} &=7Hz \end{aligned}
Then minimum sampling frequency satisfying the nyquist criterion is 7*2=14Hz.
There are 10 questions to complete.